On a bounded feasible region R, the objective function \(Z = 2x + 3y\) attains its maximum value 30 only at the point (6, 6), and its minimum value 6 only at the point (3, 0). Consider the following statements:
(A) The maximum value of \(4x + 6y\) on R is 60.
(B) The minimum value of \(-2x - 3y\) on R is \(-6\).
(C) The maximum value of \(-2x - 3y\) on R is \(-6\), attained at (3, 0).
(D) Every point (x, y) of R satisfies \(6 \le 2x + 3y \le 30\).
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A), (C) and (D) only
For every point of R,
\(6 \le 2x + 3y \le 30\)
Multiplying by 2 keeps the order:
\(12 \le 4x + 6y \le 60\)
Multiplying by \(-1\) reverses the order, so the maximum of Z becomes the minimum of \(-Z\):
\(-30 \le -2x - 3y \le -6\)
So on R: \(4x + 6y\) has maximum 60, and \(-2x - 3y\) has minimum \(-30\) at (6, 6) and maximum \(-6\) at (3, 0).
Checking the statements
(A) The maximum value of \(4x + 6y\) on R is 60. — Correct. It is \(2 \times 30\), at (6, 6).
(B) The minimum value of \(-2x - 3y\) on R is \(-6\). — Incorrect. The minimum is \(-30\), at (6, 6).
(C) The maximum value of \(-2x - 3y\) on R is \(-6\), attained at (3, 0). — Correct. It is the negative of the minimum of Z, at the same point.
(D) Every point (x, y) of R satisfies \(6 \le 2x + 3y \le 30\). — Correct. Z lies between its minimum 6 and maximum 30.